129. Sum Root to Leaf Numbers

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Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.

An example is the root-to-leaf path 1->2->3 which represents the number 123.

Find the total sum of all root-to-leaf numbers.

Note: A leaf is a node with no children.

Example:

Input: [1,2,3]
    1
   / \
  2   3
Output: 25
Explanation:
The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.
Therefore, sum = 12 + 13 = 25.

Example 2:

Input: [4,9,0,5,1]
    4
   / \
  9   0
 / \
5   1
Output: 1026
Explanation:
The root-to-leaf path 4->9->5 represents the number 495.
The root-to-leaf path 4->9->1 represents the number 491.
The root-to-leaf path 4->0 represents the number 40.
Therefore, sum = 495 + 491 + 40 = 1026.

Solution

也是用BackTracking的方法解，比較要注意的是sum因為是call by value的傳入helper所以沒辦法像之前物件那樣被更新到。我這裡用global parameter的方式去改。

    int sum = 0;
    public int sumNumbers(TreeNode root) {
        if(root == null) return 0;
        helper(root, 0);
        return sum;
    }
    
    public void helper(TreeNode node, int temp){
        
        if(node.right == null && node.left == null){
            temp = temp*10 + node.val;
            sum += temp;
            temp = (temp - node.val)/ 10;
        }
        if(node.right != null){
            temp = temp*10 + node.val;
            helper(node.right, temp);
            temp = (temp - node.val)/ 10;
        }
        if(node.left != null){
            temp = temp*10 + node.val;
            helper(node.left, temp);
            temp = (temp - node.val)/ 10;
        }
    }